类型的资源（4）（mysql链接）注意：类stdClass的对象无法转换为int

Question

我想通过添加所有progressDone并将其除以完成的进度数来计算progressDone的百分比。它不工作，我也得到错误

  

类型的资源（4）（mysql链接）
  注意：类stdClass的对象无法转换为int

$query = "SELECT SUM(progressDone) AS totall, COUNT(progressDone) AS done FROM  progress, progressType WHERE progress.projectID='$projectID' AND progress.progressTypeID='1'";
$result = mysql_query($query, $db)
or die("Query failed: ".mysql_error()." Actual query: ".$query);
$online = mysql_fetch_object($result);
$total = done; //done from the sql statement 
$current = $online;
$percent = (($current/$total) * 100);
$percentNot = 100 - $percent;

Answer 1

你的问题在这里

$online = mysql_fetch_object($result);
$total = done; //done from the sql statement 
$current = $online;

mysql_fetch_object返回一个对象。你不能像这样引用列。您需要将列作为对象的成员调用

$total = $online->done;
$current = $online->totall;